sin ^4 frac{pi}{8}+sin ^4 frac{3 pi}{8}+sin ^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We have the expression: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ Since $\sin(\pi - 	he

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) We have the expression: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$ Since $\sin(\pi - 	heta) = \sin 	heta$,we have $\sin \frac{5 \pi}{8} = \sin \frac{3 \pi}{8}$ and $\sin \frac{7 \pi}{8} = \sin \frac{\pi}{8}$. Thus,$S = 2 \left[ \sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} ight]$. Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get $\sin^4 	heta = \left( \frac{1 - \cos 2	heta}{2} ight)^2$. $S = 2 \left[ \left( \frac{1 - \cos(\pi/4)}{2} ight)^2 + \left( \frac{1 - \cos(3\pi/4)}{2} ight)^2 ight]$ $S = 2 \left[ \left( \frac{1 - 1/\sqrt{2}}{2} ight)^2 + \left( \frac{1 - (-1/\sqrt{2})}{2} ight)^2 ight]$ $S = \frac{2}{4} \left[ (1 - 1/\sqrt{2})^2 + (1 + 1/\sqrt{2})^2 ight] = \frac{1}{2} \left[ (1 + 1/2 - \sqrt{2}) + (1 + 1/2 + \sqrt{2}) ight]$ $S = \frac{1}{2} \left[ 3 ight] = \frac{3}{2}$.

$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8} = ?$

Similar Questions

$\sin 36^\circ \sin 72^\circ \sin 108^\circ \sin 144^\circ = $

If $a \tan \alpha + b \tan \beta = (a + b) \tan \left( \frac{\alpha + \beta}{2} \right)$ and $\alpha - \beta \neq 2n\pi$,then $\frac{\cos \beta}{\cos \alpha} = $

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

The value of $\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} = $

The graph of the function $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module